A stapled sequence is defined as a set of consecutive positive integers such that no one of them is relatively prime with all of the others. When I first came across this definition, I thought it wouldn't be that difficult to find an $n\in \mathbb N$ such that every set $A=\{i\in\mathbb N: j≤i≤j+n-1\}$ for some $j\in \mathbb N$ is a stapled sequence.
It turned out, however, that it wasn't that trivial. I first tried to use the "identity" $$\text{gdc}(a,b)=\text{gcd}(a-bt,b)\;\;\forall a,b,t\in\mathbb N$$but it took me nowhere.
I then found an article which seemed to help; in the abstract of this article they assert, that there's a very simple proof to show that no stapled sequence containing $N$ consecutive positive intgers exist for any $N < 17$.
Nevertheless, I've tried to read the article and I don't find their proof trivial; I don't even understand it!
(It might be worth to tell that I'm a secondary school student interested in maths).
Could someone help me understand their proof or provide a own proof?
I agree that the notations in the article are a little bit confusing and overkill if you're only interested in the result that there are no stapled sequences of length $\lt 17$. So here goes my simplified (but longer, because there are more details and explanations) presentation.
I say that two integers are friends if they are distinct and not coprime, and that they are $p$-friends (for a prime $p$) if they are distinct and both divisible by $p$. Thus, two integers are friends iff they are $p$-friends for some prime $p$, and a stapled sequence is a sequence in which everyone has a friend.
Lemma 3.6 (of the paper) If there is a stapled sequence of length $2N$, then
(a) there is also a stapled sequence of length $2N+1$.
(b) there is also a stapled sequence of length $2N-1$.
Proof of lemma 3.6 Let $S=(x+1,x+2,\ldots,x+2N)$ be a stapled sequence of length $2N$.
(a) Suppose first that $x$ is odd. Then I claim that $(x+1,x+2,\ldots,x+2N+1)$ is also stapled. Indeed, everyone except perhaps $x+2N+1$ already has a friend, and $x+2N+1$ is even and therefore a $2$-friend to $x+1$. A mirror argument shows that $(x,x+1,\ldots,x+2N)$ is stapled when $x$ is even.
(b) Suppose first that $x$ is odd. Then I claim that the follwing set is also stapled : $T=(x+1,x+2,\ldots,x+2N-1)$. To show this, we take a $y$ in $T$ and show that it has a friend in $T$. By hypothesis, $y$ has friend $z$ in $S$. If $z\neq x+2N$, then $z$ is already in $T$ and we are done. If $z=x+2N$, there is a prime $p$ that divides both $x+2N$ and $z$. If $p<N$, either $y-p$ or $y+p$ is in $T$ and is a $p$-friend of $y$, and we are done also. Finally, if $p\geq N$, then there are only two integers divisible by $p$ in $S$, namely $x+2N$ and $x+2N-p$. So $y=x+2N-p$ is even and has $2$-friends inside $T$, which finishes the proof that $T$ is stapled. A mirror argument shows that $(x+2,x+3,\ldots,x+2N)$ is stapled when $x$ is even.
Lemma 3.7 (of the paper) If there is a stapled sequence of length $p$ where $p$ is an odd prime, then there is also a stapled sequence of length $p+1$.
Proof of lemma 3.7 Let $S=(x+1,x+2,\ldots,x+p)$ be a stapled sequence of length $p$. By the Chinese remainder theorem, there is an $x'$ such that $x'$ is congruent to $x$ modulo every prime $<p$, and such that $x'+p+1$ is divisible by $p$. Then $x'+1$ and $x'+p+1$ are $p$-friends, and friends in $S=(x+1,x+2,\ldots,x+p)$ stay friends when translated by $x'-x$. Thus $S'=(x'+1,x'+2,\ldots,x'+p+1)$ is a stapled sequence.
Taking the contrapositive statements of lemmata 3.6 and 3.7, we obtain the following :
Corollary. Let $Z$ be the set of all integers $z$ such that there are no stapled sequences of length $z$. Then :
(R1) If $z\in Z$ is odd, then $z-1\in Z$ also.
(R2) If $z\in Z$ is such that $z-1$ is prime, then $z-1\in Z$ also.
(R3) If $z\in Z$ is odd, then $z+1\in Z$ also.
In particular, applying rules R1 and R2 alternatively one sees that if $15\in Z$, then all of $14,13,12,11,10$ are in $Z$ ; also $16\in Z$ by R3. Similarly, if $9\in Z$ then all of $8,7,6,5,4,3,2$ are in $Z$.
So, to show that there are no stapled sequences in length $\leq 16$, it suffices to show that there are no stapled sequences in length $9$ or $15$.
Theroem 3.8 (of the paper) There are no stapled sequences of length $\leq 16$.
Proof of theorem 3.8 from lemma. As noted above, it suffices to show that
(a) There are no stapled sequences in length $9$.
(b) There are no stapled sequences in length $15$.
(a) : Suppose that $S=(x+1,x+2, \ldots,x+9)$ is stapled. The set $Y=\lbrace x+1,x+3,x+5,x+7,x+9\rbrace$ contains at most two multiples of $3$, at most one multiple of $5$ and at most one multiple of $7$. So some $y\in Y$ is not divisible by any of $3,5,7$. We know that $y$ has a friend $z$ in $S$, so they are both divisible by a prime $p$. The only possibility is then $p=2$, so $x$ is odd. Next, consider $U=\lbrace x+2,x+4,x+6,x+8\rbrace$. This set contains at most one multiple of $5$ and at most one multiple of $7$, so there are two elements $u_1,u_2 \in U$ not divisible by $5$ or $7$. We know that $u_i(1\leq i \leq 2)$ has a friend $v_i$ in $S$, so they are both divisible by a prime $p_i$. The only possibility is then $p_i=3$ (for both values of $i$). So both $u_1$ and $u_2$ are divisible by $3$ ; but the only two elements congruent to each other modulo $3$ in $U$ are $x+2$ and $x+8$. It follows that $x\equiv 1[\mod 3]$. Combining the two congruences that we now have on $x$, we see that $x$ is of the form $6q+1$. But then $x+4=6q+5$ and $x+6=6q+7$, and one of those two numbers is not divisible by any of $2,3,5$, and therefore cannot have any friends in $S$.
(b) : Suppose that $S=(x+1,x+2, \ldots,x+15)$ is stapled.
Each element of $A(x)=\lbrace x+3,x+4,\ldots,x+13\rbrace$ has a friend in $S$ which is at at most $12$ of distance to it, and so must be divisible by a prime $\leq 12$.
Each element of $B(x)=\lbrace x+5,x+6,\ldots,x+11\rbrace$ has a friend in $S$ which is at at most $10$ of distance to it, and so must be divisible by a prime $\leq 10$.
Next, let $C(x)$ be set of all elements of $A(x)$ not divisible by $2$ or $3$. The content of $C(x)$ depends of the value of $x$ modulo $6$ :
$$ \begin{array}{|l|c|} \hline x \ (mod\ 6) & C(x) \\ \hline 0 & \lbrace x+5,x+7,x+11,x+13\rbrace \\ \hline 1 & \lbrace x+4,x+6,x+10,x+12\rbrace \\ \hline 2 & \lbrace x+3,x+5,x+9,x+11\rbrace \\ \hline 3 & \lbrace x+4,x+8,x+10\rbrace \\ \hline 4 & \lbrace x+3,x+7,x+9,x+13\rbrace \\ \hline 5 & \lbrace x+6,x+8,x+12,x+14\rbrace \\ \hline \end{array} $$
We see then that $x \ (mod\ 6)$ is not $3$ or $4$, then $C(x)$ is a set of four elements that are not congruent to each other modulo $5,7$ or $11$, so that one of them is not divisible by $5,7$ or $11$, which is impossible. So we can assume that $x \ (mod\ 6)$ is $3$ or $4$. Next, let $D(x)$ be set of all elements of $C(x)$ not divisible by $2$ or $3$. We have :
$$ \begin{array}{|l|c|} \hline x \ (mod\ 6) & D(x) \\ \hline 3 & \lbrace x+8,x+10\rbrace \\ \hline 4 & \lbrace x+5,x+9,x+11\rbrace \\ \hline \end{array} $$
Reasoning as before, we see that $x \equiv 4 (mod\ 6)$ is impossible. So we must have $x \equiv 3 (mod\ 6)$, and further, one of $x+8,x+10$ is divisible by $5$, and the other is divisible by $7$.
But then, in $C(x)=\lbrace x+4,x+8,x+10\rbrace$, the remaining $x+4$ is not divisible by $5$ or $7$ and so cannot have any friends in $S$. This finishes the proof.